Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 26

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 4,\,\,x = - 2,\,\,x = 0,\,\,x = 3,\,\,\,x = 4 \cr & \left( b \right)f\left( { - 4} \right) = \frac{{142}}{3}{\text{ Is the absolute maximum}} \cr & \,\,\,\,\,\,f\left( 3 \right) = - \frac{{23}}{4}{\text{is the absolute }}\left( {{\text{and local}}} \right)\,{\text{minimum}} \cr & \,\,\,\,\,\,\,f\left( { - 2} \right) = \frac{{14}}{3}{\text{ is a local minimum}} \cr & \,\,\,\,\,\,\,f\left( 0 \right) = 10{\text{ is a local maximum}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^4}}}{4} - \frac{{{x^3}}}{3} - 3{x^2} + 10,{\text{ on the interval }}\left[ { - 4,4} \right] \cr & \cr & {\text{Because }}f{\text{ is a polynomial}}{\text{, its derivative exists everywhere}} \cr & f'\left( x \right) = \frac{{4{x^3}}}{4} - \frac{{3{x^2}}}{3} - 6x \cr & f'\left( x \right) = {x^3} - {x^2} - 6x \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & {x^3} - {x^2} - 6x = 0 \cr & x\left( {{x^2} - x - 6} \right) = 0 \cr & x\left( {x - 3} \right)\left( {x + 2} \right) = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = 0,\,\,x = - 2\,{\text{ and }}x = 3 \cr & {\text{All of them lies in the given interval }}\left[ { - 4,4} \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ { - 4, - 2,0,3,4} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 4} \right) = \frac{{{{\left( { - 4} \right)}^4}}}{4} - \frac{{{{\left( { - 4} \right)}^3}}}{3} - 3{\left( { - 4} \right)^2} + 10 = \frac{{142}}{3} \cr & f\left( { - 2} \right) = \frac{{{{\left( { - 2} \right)}^4}}}{4} - \frac{{{{\left( { - 2} \right)}^3}}}{3} - 3{\left( { - 2} \right)^2} + 10 = \frac{{14}}{3} \cr & f\left( 0 \right) = \frac{{{{\left( 0 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^3}}}{3} - 3{\left( 0 \right)^2} + 10 = 10 \cr & f\left( 3 \right) = \frac{{{{\left( 3 \right)}^4}}}{4} - \frac{{{{\left( 3 \right)}^3}}}{3} - 3{\left( 3 \right)^2} + 10 = - \frac{{23}}{4} \cr & f\left( 4 \right) = \frac{{{{\left( 4 \right)}^4}}}{4} - \frac{{{{\left( 4 \right)}^3}}}{3} - 3{\left( 4 \right)^2} + 10 = \frac{{14}}{3} \cr & \cr & {\text{The largest of these function values is }}f\left( { - 4} \right) = \frac{{142}}{3},{\text{ which is the}} \cr & {\text{absolute maximum on the interval }}\left[ { - 4,4} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( 3 \right) = - \frac{{23}}{4}{\text{which is the absolute}} \cr & \left( {{\text{and local}}} \right){\text{minimum on the interval }}\left[ { - 4,4} \right]. \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$